$3.1.14^*.$ A weightless rod of length $L > l$ with masses $m$ at its ends hangs on threads of length $h$ located at a distance $l$ from each other. The rod is horizontal. Show that when the rod is rotated about its vertical axis of symmetry by a small angle $\varphi$, the torque of the forces acting on the masses is proportional to $\varphi$, and the change in the potential energy of the masses is proportional to $\varphi^2$. Find the maximum angular velocity of the rod if it is released after being initially rotated by an angle $\varphi_0$.
### Solution
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Let's analyze the geometry of the displacement when the rod is rotated by an angle $\varphi$. The point of attachment of the thread moves along an arc. The horizontal displacement of the thread from its vertical position is given by the chord $A'C$.
From the side view, we can express this displacement as:
$$ A'C = h \sin\alpha$$
where $\alpha$ is the angle of deviation of the thread from the vertical.
Considering the top view, let the center of rotation be $O'$. The distance to the thread attachment point is $O'C = O'A' = \frac{l}{2}$.
Using the Law of Sines in the isosceles triangle $\triangle O'CA'$:
Next, let's find the torque $\vec{M}$ caused by the tension force $\vec{T}$. Using cylindrical coordinates, the position vector of the thread attachment point is $\vec{r} = \frac{l}{2}\vec{e}_r$. The tension force can be decomposed as:
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$$\vec{T} = -T \sin\alpha\vec{e}_\varphi + T \cos\alpha\vec{e}_z $$
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Next, let's find the restoring torque torque $\vec{M} = [\vec{r}\times\vec{T}]$ caused by the tension force $T$.
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The torque is $\vec{M} = [\vec{r}\times\vec{T}]$. The vertical component $M_z$, which twists the rod back toequilibrium, is:
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$$M_z = -\frac{l}{2} T \sin\alpha$$
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The tension force has a horizontal component that pulls the rod back towards its equilibrium position. This horizontal force is:
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$$F_h = T \sin\alpha$$
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For a small angle $\varphi$, this horizontal force acts approximately perpendicular to the rod. The distance from the axis of rotation to the thread attachment point (the lever arm) is $l/2$. Therefore, the restoring torque is simply the product of the force and the lever arm:
$$ z = h \left( 1 - \left[ 1 - \left( \frac{l \sin\varphi}{2h \cos\left(\frac{\varphi}{2}\right)}\right)^2 \right]^{1/2}\right) $$
For small angles $\varphi\to 0$, applying the approximation $(1+x)^a \approx 1 + ax$, we obtain:
$$ z \approx h \left( 1 - \left[ 1 - \left( \frac{l\varphi}{2h}\right)^2 \right]^{1/2}\right) \approx h \left( 1 - 1 + \frac{1}{2}\left(\frac{l}{2h}\right)^2 \varphi^2 \right) = \frac{l^2}{8h}\varphi^2 $$
This proves that the change in potential energy is proportional to the square of the angle: $U \sim\varphi^2$.
Finally, to find the maximum angular velocity $\omega$, we apply the law of conservation of energy. The maximum kinetic energy at the equilibrium position equals the initial potential energy at the maximum deflection angle $\varphi_0$:
$$\frac{mv^2}{2} = mgz \implies v = \sqrt{2gz}$$
The linear velocity of the masses at the ends of the rod is related to the angular velocity by $v = \omega\frac{L}{2}$. Therefore:
$$\omega = \frac{2v}{L} = \frac{2\sqrt{2gz}}{L}$$
Substituting $z = \frac{l^2}{8h}\varphi_0^2$ into the equation:
$3.1.14^*.$ A weightless rod of length $L > l$ with masses $m$ at its ends hangs on threads of length $h$ located at a distance $l$ from each other. The rod is horizontal. Show that when the rod is rotated about its vertical axis of symmetry by a small angle $\varphi$, the torque of the forces acting on the masses is proportional to $\varphi$, and the change in the potential energy of the masses is proportional to $\varphi^2$. Find the maximum angular velocity of the rod if it is released after being initially rotated by an angle $\varphi_0$.
$3.1.14^*.$ A weightless rod of length $L > l$ with masses $m$ at its ends hangs on threads of length $h$ located at a distance $l$ from each other. The rod is horizontal. Show that when the rod is rotated about its vertical axis of symmetry by a small angle $\varphi$, the torque of the forces acting on the masses is proportional to $\varphi$, and the change in the potential energy of the masses is proportional to $\varphi^2$. Find the maximum angular velocity of the rod if it is released after being initially rotated by an angle $\varphi_0$.
### Solution
### Solution
Let's analyze the geometry of the displacement when the rod is rotated by an angle $\varphi$. The point of attachment of the thread moves along an arc. The horizontal displacement of the thread from its vertical position is given by the chord $A'C$.
Let's analyze the geometry of the displacement when the rod is rotated by an angle $\varphi$. The point of attachment of the thread moves along an arc. The horizontal displacement of the thread from its vertical position is given by the chord $A'C$.
From the side view, we can express this displacement as:
From the side view, we can express this displacement as:
$$ A'C = h \sin\alpha$$
$$ A'C = h \sin\alpha$$
where $\alpha$ is the angle of deviation of the thread from the vertical.
where $\alpha$ is the angle of deviation of the thread from the vertical.
Considering the top view, let the center of rotation be $O'$. The distance to the thread attachment point is $O'C = O'A' = \frac{l}{2}$.
Considering the top view, let the center of rotation be $O'$. The distance to the thread attachment point is $O'C = O'A' = \frac{l}{2}$.
Using the Law of Sines in the isosceles triangle $\triangle O'CA'$:
Using the Law of Sines in the isosceles triangle $\triangle O'CA'$:
Next, let's find the torque $\vec{M}$ caused by the tension force $\vec{T}$. Using cylindrical coordinates, the position vector of the thread attachment point is $\vec{r} = \frac{l}{2}\vec{e}_r$. The tension force can be decomposed as:
Next, let's find the restoring torque torque $\vec{M} = [\vec{r}\times\vec{T}]$ caused by the tension force $T$.
$$\vec{T} = -T \sin\alpha\vec{e}_\varphi + T \cos\alpha\vec{e}_z $$
The torque is $\vec{M} = [\vec{r}\times\vec{T}]$. The vertical component $M_z$, which twists the rod back toequilibrium, is:
The tension force has a horizontal component that pulls the rod back towards its equilibrium position. This horizontal force is:
$$M_z = -\frac{l}{2} T \sin\alpha$$
$$F_h = T \sin\alpha$$
For a small angle $\varphi$, this horizontal force acts approximately perpendicular to the rod. The distance from the axis of rotation to the thread attachment point (the lever arm) is $l/2$. Therefore, the restoring torque is simply the product of the force and the lever arm:
$$ z = h \left( 1 - \left[ 1 - \left( \frac{l \sin\varphi}{2h \cos\left(\frac{\varphi}{2}\right)}\right)^2 \right]^{1/2}\right) $$
$$ z = h \left( 1 - \left[ 1 - \left( \frac{l \sin\varphi}{2h \cos\left(\frac{\varphi}{2}\right)}\right)^2 \right]^{1/2}\right) $$
For small angles $\varphi\to 0$, applying the approximation $(1+x)^a \approx 1 + ax$, we obtain:
For small angles $\varphi\to 0$, applying the approximation $(1+x)^a \approx 1 + ax$, we obtain:
$$ z \approx h \left( 1 - \left[ 1 - \left( \frac{l\varphi}{2h}\right)^2 \right]^{1/2}\right) \approx h \left( 1 - 1 + \frac{1}{2}\left(\frac{l}{2h}\right)^2 \varphi^2 \right) = \frac{l^2}{8h}\varphi^2 $$
$$ z \approx h \left( 1 - \left[ 1 - \left( \frac{l\varphi}{2h}\right)^2 \right]^{1/2}\right) \approx h \left( 1 - 1 + \frac{1}{2}\left(\frac{l}{2h}\right)^2 \varphi^2 \right) = \frac{l^2}{8h}\varphi^2 $$
This proves that the change in potential energy is proportional to the square of the angle: $U \sim\varphi^2$.
This proves that the change in potential energy is proportional to the square of the angle: $U \sim\varphi^2$.
Finally, to find the maximum angular velocity $\omega$, we apply the law of conservation of energy. The maximum kinetic energy at the equilibrium position equals the initial potential energy at the maximum deflection angle $\varphi_0$:
Finally, to find the maximum angular velocity $\omega$, we apply the law of conservation of energy. The maximum kinetic energy at the equilibrium position equals the initial potential energy at the maximum deflection angle $\varphi_0$:
$$\frac{mv^2}{2} = mgz \implies v = \sqrt{2gz}$$
$$\frac{mv^2}{2} = mgz \implies v = \sqrt{2gz}$$
The linear velocity of the masses at the ends of the rod is related to the angular velocity by $v = \omega\frac{L}{2}$. Therefore:
The linear velocity of the masses at the ends of the rod is related to the angular velocity by $v = \omega\frac{L}{2}$. Therefore:
$$\omega = \frac{2v}{L} = \frac{2\sqrt{2gz}}{L}$$
$$\omega = \frac{2v}{L} = \frac{2\sqrt{2gz}}{L}$$
Substituting $z = \frac{l^2}{8h}\varphi_0^2$ into the equation:
Substituting $z = \frac{l^2}{8h}\varphi_0^2$ into the equation:
